Derivatives involving both product/quotient rule

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Dec 6th 2009, 09:33 AM
Archduke01

Derivatives involving both product/quotient rule

Find f'(x) for y = (x+5 / x+1) (2x+1)

I don't know how to proceed here. Do I use the quotient rule inside the first brackets? Then what?
Dec 6th 2009, 09:43 AM
lvleph

First thing you can do is use the product rule
$y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot<br /> \frac{x+5}{x+1}$ .
From here we can take the derivative using the quotient rule on $\frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}$ .
Combine this with the previous result
$y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot<br /> \frac{x+5}{x+1}$ .
Now simplify.
Dec 6th 2009, 09:51 AM
Archduke01

Quote:

Originally Posted by lvleph

First thing you can do is use the product rule
$y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot<br /> \frac{x+5}{x+1}$ .
From here we can take the derivative using the quotient rule on $\frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}$ .
Combine this with the previous result
$y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot<br /> \frac{x+5}{x+1}$ .
Now simplify.

You don't end up with $2(x^2 + 2x + 3) / (x+1)^2$ do you?
Dec 6th 2009, 10:10 AM
lvleph

Yes, that is what www.wolframalpha.com gives.

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